# Set up, and evaluate the integral that represents the surface area of rotation of this cardioid about the y-axis.

#### Elix

##### New member
• MarkFL

#### MarkFL

##### Super Moderator
Staff member
I don't see any errors in your work so far. Let's write (after some simplification):

$$\displaystyle A=2\sqrt{2}\pi\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(\theta)(\sin(\theta)+1)^{\frac{3}{2}}\,d\theta$$

Now it appears the integration should be straightforward. #### MarkFL

##### Super Moderator
Staff member
To follow up, let's let:

$$\displaystyle u=\sin(\theta)+1\implies du=\cos(\theta)\,d\theta$$

And our definite integral becomes:

$$\displaystyle A=2\sqrt{2}\pi\int_0^2 u^{\frac{3}{2}}\,du=2\sqrt{2}\pi\left(\frac{2}{5}2^{\frac{5}{2}}\right)=\frac{32\pi}{5}$$

#### Elix

##### New member
To follow up, let's let:

$$\displaystyle u=\sin(\theta)+1\implies du=\cos(\theta)\,d\theta$$

And our definite integral becomes:

$$\displaystyle A=2\sqrt{2}\pi\int_0^2 u^{\frac{3}{2}}\,du=2\sqrt{2}\pi\left(\frac{2}{5}2^{\frac{5}{2}}\right)=\frac{32\pi}{5}$$
Thanks a bunch! I struggled to connect how you got to that simplification at first but figured it out. I really appreciate your help with this and my previous question!

• MarkFL